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A wheel of moment of inertia 0.10 is rotating about a shaft at an angular speed of 160rev/min.A second wheel is set into rotation at 300 rev/min and is coupled to the same shaft so that both the wheels finally rotate with the common angular velocity of 200 rev/min.Find the moment of inertis of second wheel
For this purposeuse the law of conservation of angular momentum
I1(omega)1(Angular mom. of first wheel )+I2(omega)2 (Angular mom. of seconn d wheel )=(I1+I2)(omega)f (Angular mom. of combined system wheel )
(0.1)(320pi rad/min) + I2(600pi rad/min)=(0.1+I2)(400pi rad/min)
Solve the above equation and u will get the value of I2
I1 and I2 are moments of inertia
w1=160,w2=300,w3=200,
I1*w1 + I2*w2=(I1 + I2)*w3
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